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\(\bullet\) A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one-fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

Short Answer

Expert verified
(a) The volume increases as pressure decreases. (b) The volume change would be twice as great. (c) The volume change in lead is four times greater than in gold.

Step by step solution

01

Understand the Concept of Volume Change

The volume of a solid object changes as the pressure applied on it changes. This phenomenon is governed by the object's bulk modulus (K), which describes an object's resistance to uniform pressure. The change in volume \( \Delta V \) relative to the original volume \( V \) can be calculated using the formula: \[ \frac{\Delta V}{V} = - \frac{\Delta P}{K} \] where \( \Delta P \) is the change in pressure.
02

Analyze Volume Change from Ship to Ocean Surface

As the gold bar is brought up from the ship to the surface, the pressure decreases. This results in an increase in its volume, since pressure and volume are inversely related when bulk modulus is considered. The exact change in volume can be determined if \( \Delta P \) and \( K \) (bulk modulus of gold) are known.
03

Examine Volume Change with Depth Doubling

The pressure exerted on an object underwater increases linearly with depth. If the ship were twice as deep, the pressure \( \Delta P \) would be twice as much. Since volume change \( \Delta V \) is directly proportional to \( \Delta P \) (from the formula in Step 1), the volume change would also be twice as great.
04

Calculate the Ratio of Volume Changes

The bulk modulus of lead is one-fourth that of gold. Hence, the formula \( \frac{\Delta V}{V} = - \frac{\Delta P}{K} \) implies that the volume change for the same pressure difference is inversely proportional to the bulk modulus. Thus, the volume change in lead is four times greater than that in gold for the same pressure change.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Change
When an object is subjected to pressure, its volume can change. This is particularly true for solid materials like metals. The volume of a solid object is inversely related to the pressure applied to it. That means when pressure decreases, volume increases, and vice versa.

The bulk modulus (\( K \)) of the material tells us how much a material resists volume change under pressure. The formula to calculate the fractional change in volume is:
  • \( \frac{\Delta V}{V} = - \frac{\Delta P}{K} \)
Here, \( \Delta V \) is the change in volume, \( V \) is the original volume, and\( \Delta P \) is the change in pressure. Understanding this relationship helps us predict how the volume of an object will react when brought from a high-pressure environment to a lower-pressure one, such as a gold bar being lifted from the depths of the ocean to the surface.
Pressure Difference
Pressure difference refers to the change in pressure exerted on an object when it moves from one environment to another. In the context of an object being submerged underwater, pressure increases with depth. Therefore, when an object like a gold bar is lifted from deep underwater, it experiences a decrease in pressure as it ascends.

This pressure change (\( \Delta P \)) is crucial, as it influences how much the object's volume will change. Mathematically, since pressure and volume change are directly proportional in these situations, if the original pressure difference is doubled, the corresponding volume change also doubles. Understanding this concept is key, as it provides insights into how material properties can be manipulated by altering environmental pressures.
Gold
Gold is a dense, lustrous metal that exhibits unique properties making it highly resistant to corrosion and tarnish. It also has a high bulk modulus, which means that it is relatively resistant to changes in volume when pressure is applied. This is why gold will have a smaller volume change compared to materials with lower bulk modulus values, such as lead.

The bulk modulus of gold is significant when calculating how it will react to pressure changes. Since the formula \( \frac{\Delta V}{V} = - \frac{\Delta P}{K} \) shows that volume change is inversely proportional to the bulk modulus, having a high value of \( K \) implies that the material undergoes less volume change for the same pressure difference. This innate property of gold makes it stable concerning structural shape when used in a variety of environments.
Lead
Lead is a heavy and soft metal known for its high density and low bulk modulus. The bulk modulus of lead is one-fourth that of gold, which indicates its relatively high susceptibility to volume changes under pressure.

For instance, when the same pressure difference (\( \Delta P \)) is applied to both a gold and a lead bar, the lead bar will undergo a greater volume change. Specifically, the volume change in lead will be four times greater than that in gold. This outcome is obtained because the formula \( \frac{\Delta V}{V} = - \frac{\Delta P}{K} \) demonstrates that a lower bulk modulus results in a larger volume alteration. Understanding the characteristics of lead under different conditions allows engineers and scientists to predict how such materials will behave in various applications.

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Most popular questions from this chapter

\(\bullet\) Achilles tendon. The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is \(78.1 \mathrm{mm}^{2},\) its average length is \(25 \mathrm{cm},\) and its average Young's modulus is 1474 MPa. (a) How much tensile stress is required to stretch this muscle by 5.0\(\%\) of its length? (b) If we model the tendon as a spring, what is its force constant? (c) If a 75 kg sprinter exerts a force of 13 times his weight on his Achilles tendon, by how much will it stretch?

\(\bullet\) (a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern having the same frequency of the note that is sung. If someone sings the note \(B\) flat that has a frequency of 466 \(\mathrm{Hz}\) , how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angular frequency of the cords? (b) Hearing. When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that typical humans can hear has a period of 50.0\(\mu \mathrm{s} .\) What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) Vision. When light having vibrations with angular frequency ranging from \(2.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) to \(4.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) Ultrasound. High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect a small objects such as tumors, a frequency of around 5.0 \(\mathrm{MHz}\) is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?

\(\bullet\) A mass of 0.20 \(\mathrm{kg}\) on the end of a spring oscillates with a period of 0.45 s and an amplitude of 0.15 \(\mathrm{m}\) . Find (a) the velocity when it passes the equilibrium point, (b) the total energy of the system, and (c) the equation describing the motion of the mass, assuming that \(x\) was a maximum at time \(t=0\) .

\(\bullet\) \(\bullet\) The effect of jogging on the knees. High-impact activities such as jogging can cause considerable damage to the cartilage at the knee joints. Peak loads on each knee can be eight times body weight during jogging. The bones at the knee are separated by cartilage called the medial and lateral meniscus. Although it varies considerably, the force at impact acts over approximately 10 \(\mathrm{cm}^{2}\) of this cartilage. Human cartilage has a Young's modulus of about 24 MPa (although that also varies). (a) By what percent does the peak load impact of jogging compress the knee cartilage of a 75 kg person? (b) What would be the percentage for a lower-impact activity, such as power walking, for which the peak load is about four times body weight?

\(\bullet\) A 1.35 \(\mathrm{kg}\) object is attached to a horizontal spring of force constant 2.5 \(\mathrm{N} / \mathrm{cm}\) and is started oscillating by pulling it 6.0 \(\mathrm{cm}\) from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.5 \(\mathrm{cm}\) . (a) How much energy has this system lost to damping during these eight cycles? (b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

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